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Proof step 10 from proof gcdexh (
previous step
)
∀ m0, n0. m0 < n → n0 ≤ m0 → (∃ m1, n1. m0 * m1 + gcd(m0, n0) = n0 * n1 ∨ n0 * n1 + gcd(m0, n0) = m0 * m1) ⊦ n = 0, ∃ m, n0. m + 1 = n0 ∨ n0 + 1 = m
User applied rule exists right
And closed the goal!
Rule argument
pos: right 2 terms: 0, 1 abort: F discard: T computed: F
Additional information
Used terms: 0, 1